First slide

Introduction to limits

Question

Let $m$ and $n$ be two positive integers greater than $1 .$ If $lim_{α \to 0} (\frac{e^{\cos (α^{n})} - e}{α^{'''}}) = - \frac{c}{2}$ , than the value of $\frac{m}{n},$ is

Moderate

A

2
B

4
C

6
D

9

Solution

We have,
$\begin{array}{l} lim_{α \to 0} (\frac{e^{\cos (α^{n})} - e}{α^{m}}) = - \frac{e}{2} \\ \Rightarrow - e lim_{α \to 0} (\frac{e^{\cos (α^{n})} - 1}{\cos (α^{n}) - 1}) (\frac{1 - \cos (α^{n})}{α^{m}}) = - \frac{e}{2} \end{array}$
$\Rightarrow - e lim_{α \to 0} (\frac{e^{\cos (α^{n}) - 1} - 1}{\cos (α^{n}) - 1}) (\frac{1 - \cos (α^{n})}{α^{2 n}}) α^{2 n - m} = - \frac{e}{2}$
$\begin{array}{r} \Rightarrow - e lim_{α \to 0} \frac{e^{\cos (u^{n}) - 1} - 1}{\cos (α^{n}) - 1} \times lim_{α \to 0} (\frac{1 - \cos (α^{n})}{α^{2 n}}) \times lim_{α \to 0} α^{2 n - m} \\ = - \frac{e}{2} \end{array}$
$\begin{array}{l} \Rightarrow - e \times 1 \times \frac{1}{2} \times lim_{α \to 0} α^{2 n - m} = - \frac{e}{2} \\ \Rightarrow lim_{α \to 0} α^{2 n - m} = 1 \Rightarrow 2 n - 1 n = 0 \Rightarrow \frac{m}{n} = 2 \end{array}$

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